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## Calculus 2

### Unit 2: Lesson 2

Integrating using long division and completing the square# Integration using long division

AP.CALC:

FUN‑6 (EU)

, FUN‑6.D (LO)

, FUN‑6.D.3 (EK)

Here we do polynomial long division to make an integral more computable.

## Want to join the conversation?

- I was working to find the integral of 3/(2x+4) using u-sub. If I factor out the 2 from the denominator and set u = x+2 I get the following solution: 3/2 * ln(x+2). However, if I do not factor and let u = 2x+4, I get 3/2 * ln(2x +4). What am I missing? Thanks.(20 votes)
- Remember that a general antiderivative of a function (indefinite integral) always has a constant of integration c attached to it. Assuming the above integral was done correctly, there should be a c attached to both. Notice that the first solution is 3/2 * ln(x+2) +c and the second is 3/2 * ln(2x+4) + c. Now manipulate (3/2) ln(2x+4) + c to get (3/2) ln(2*(x+2) ) + c and you get (3/2) ln(2) + (3/2) ln(x+2) +c by log properties. Notice now that 3/2 * ln(2) can be absorbed into the constant of integration, because it is a real number. Thus, we get (3/2) ln(x+2) + c for both the first and second solutions.

Always remember to include the constant of integration.(38 votes)

- He did the long division as if this is something I should kinda know already is this covered in algebra or something?(10 votes)
- Hi, I need help solving a problem.

The problem is to find the Anti-Derivative of x^2 + 2x - 2 + 15/(2x+3)

The answer I came up with is : x^3/3 + x^2 - 2x + 15ln(2x+3) + C

However, the correct answer is coming up as :

:x^3/3 + x^2 - 2x + (15ln(2x+3))/2 + C

My answer is almost identical to the correct answer except for the last part.

I don't understand why the last expression is divided by 2.

Can someone explain this to me please?

Thank you.(4 votes)- Notice that in the standard integral forms, you ALWAYS have a du. This is NOT just some notation that you can ignore, it is vital to getting the correct answer. The du is the derivative of whatever you call u -- it MUST be present or you cannot use that standard derivative form.

for the`∫ (1/u) du = ln(u) + C`

form, you must have the derivative of the denominator you are calling u.

For`15/(2x+3)`

, you declared`u`

to be`2x+3`

. That is fine, but you MUST have its derivative.`u = 2x+3`

`du = 2 dx`

in other words`dx = ½ du`

So, to use this form I would need to do the following:`∫ 15 dx / (2x+3)`

← though not necessary, factor out the 15 to avoid mistakes`15∫ dx / (2x+3)`

← now replace what you have with the u and du.`15∫ (½ du) / (2x+3)`

← we do this because dx = ½du`15∫ (½ du) / u`

← no we need to make this EXACTLY match the standard form.`(15/2) ∫ du/ u`

← I now have a standard form and can integrate`(15/2) ln(u) + C`

← now back substitute`(15/2) ln(2x+3) + C`

Of course, we wouldn't typically show all of these steps, i just included them so you could get the idea.

But the big thing to understand is this: In ALL of the standard integral forms there is a du. This is the derivative the whatever you call u. This MUST be present or you cannot use this form.

For beginners, using u-substitution to make absolutely sure you have an exact match is the wisest course. Though, once you become proficient, you can probably skip the substitution and go straight to the integration. Here is the same calculation the way that I do it (this involves putting a bracket around my du, making sure I have the derivative of what I am calling u, although I won't actually use any substitutions:`∫ 15 dx / (2x+3)`

← The "u" is 2x+3, so the "du" is 2dx.`∫ 15 (½) [2 dx] / (2x+3)`

← notice that I multiplied and divided by 2, so I haven't changed anything.

`(15/2) ∫ [2 dx] / (2x+3)`

← factored out the constants that are not in the standard integral form I am using.

`= (15/2) ln(2x+3) + C`

(9 votes)

- why would he use u substitution the point where he got 2/x-1 it's just seems unnecessary(3 votes)
- If you're able to solve the antiderivative without u-sub, then great. This is meant to instruct those who just starting out with antiderivatives.(9 votes)

- https://imgur.com/a/qA1ScKo

In the above question for the integral of 1/(2x+6), if you factor out a 1/2 from the equation it becomes 1/2* integral of 1/(x+3) then doing u-sub you get 1/2*ln(x+3).

How do you know when to factor out something versus not factoring something out because the 2 answers are different?(4 votes)- It's
*technically*the same thing.

1 / 2 * ln(2x + 6) = 1 / 2 * [ln(2) + ln(x + 3)] = 1 / 2 * ln(x + 3) + C,

where C = 1 / 2 * ln 2. The two integrals have a constant difference, and are therefore technically equivalent.(3 votes)

- I am getting an extra +1/2 term compared with Sal's answer. I checked it with WolframAlpha and I get the same result as mine. Out of curiosity, I factored out from the denominator a (-2) and used u-sub (u = x - 1). I am trying to check for mistakes in my solution. Thanks for checking it out!(2 votes)
- One half is a constant, because you have a constant of integration, you can juts ignore the 1/2 as 1/2 plus C is just a different C.(4 votes)

- Can you factor out a (-2) from the denominator, changing the integral to (-1/2)*integral (x-5)/(x+1)? I get a different, but similar, answer: (-1/2)[x- 6 ln |x+1|] + C

Appreciate the clarification!(2 votes) - In the video, "Integration using long division" the fraction 4/(2x-2) is simplified to 2*(1/(x-1)) to result in 2*ln(|x-1|). However, if you leave the fraction as 2*(2/(2x-2)) the result could be integrated as 2*ln(|2x-2|) which is a different function.

1) is simplification before integration mandatory in these cases?

2) If so, how do you avoid potentially changing a function when going to a derivative and back again?(2 votes)- ln|2𝑥 − 2| = ln|(𝑥 − 1) ∙ 2| = ln|𝑥 − 1| + ln 2.

Thereby, ln|2𝑥 − 2| + 𝐶 is the same set of solutions as ln|𝑥 − 1| + 𝐶, and it doesn't really matter which representation we choose, it's just that (𝑥 − 1) is a little neater than (2𝑥 − 2), because it only involves one operation rather than two.(2 votes)

- http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s10_qp_32.pdf

I really don't understand Q10 from this link! Can someone please help me?(1 vote) - I don't know if someone asked this already, but what would you do if the denominator has a greater power than the numerator (like x^2 for the denominator and x for the numerator)(2 votes)
- it still might work to do long division, there are also cases it may not work with an equal or lower degree for the denominator as well, in which case you just want to try other ways of simplifying or rewriting the numerator and denominator.(2 votes)

## Video transcript

- [Voiceover] See if you can evaluate this integral right over here. Assuming you had a goal at it so let's work through this together. You probably realized that
some of the traditional techniques that we've already had in our tool kits don't seem to be directly applicable, u-substitution and others. The key here to realize
is we have a rational expression here where the
numerator has the same degree or higher than the denominator. In this case, the numerator and the denominator have the same degree. Whenever you see something
like that, it's probably a good idea to divide the denominator
into the numerator. That's what this rational expression could be interpreted as. X minus five, divided by
negative two x plus two. So it's a little bit of
algebraic long division. To actually divide negative two x plus two into x minus five is to
see if we can rewrite this in a way where we can
evaluate the integral. So let's do that. We're going to take x minus five. X minus five, and divide negative
two x plus two into that. Negative two x plus two. Look at the highest degree terms. How many times does
negative two x go into x? Let's going to go negative 1/2 times. Negative 1/2 times two is negative one. Negative 1/2 times negative two x is this is going to be positive x. Just like that. Now we want to subtract
this yellow expression from this blue expression,
and so let's just, I'll just take the negative
of this and then add. So I'm just going to take
the negative of it and add. We are left with negative five
plus one is negative four. You could say negative two
x plus two goes into x minus five, negative 1/2 times
with negative four left over. So we can rewrite this integral,
our original integral as, we can rewrite it as
negative 1/2 minus four over negative two x plus two, d x. Now let's see, looks like we can simplify this expression a little bit more. The numerator and the denominator, they're both divisible by two. All of these terms are divisible by two. Actually, we all have these negatives that always unnecessarily
complicate things. Let's actually divide the numerator and the denominator by negative two. What are we going to have then? We divide the numerator by negative two. If this is negative four, this is going to become positive two. Then this, if we divide negative two x by negative two, that's
just going to become x. Then two divided by negative
two is going to be minus one. So our original integral... This is just all algebra. Everything we've done so far is algebra. We just rewritten it using a little bit of algebraic long division. Our original integral has
simplified to negative 1/2, and some might argue it's not simplified but it's actually much more
useful for finding the integral. Negative 1/2 plus two
over x minus one, d x. Now, how do we evaluate this? Well, the antiderivative of negative 1/2 is
pretty straight forward. That's just going to be negative 1/2 x, plus the entire derivative
of two over x minus one. You might reduce here ahead. The derivative of x minus one is just one so you could say that the
derivative is sitting there. We can essentially use
substitution in our heads and say, "Okay, let's just
take the entire derivative, "we could say with respect
to x minus one which will be "the natural log of the
absolute value of x minus one." If all of that sounds really confusing, I'll let you do the u-substitution. If I were just trying to evaluate the integral two over x minus one, d x, I could see, okay, the derivative
of x minus one is just one so I could say u is equal to x minus one, and then d u is going to be equal to d x. This is going to be, we can
rewrite in terms of u as two, I'll just take the constant
out two times the integral of one over u, d u, which
we know as two times the natural log of the
absolute value of u plus c. In this case, we know
that u is x minus one. This is equal to two times the natural log of x minus one plus c. That's what we're going
to have right over here. So plus two times the natural log of the absolute value
of x minus one plus c. The plus c doesn't just
come from this one, this general overtaking the
integral of the whole thing. It could be some constant
because if we go the other way, we take the derivative and
the constant will go away. Let me just put the
plus c right over there, and we are done.